Problem: Simplify the following expression and state the conditions under which the simplification is valid. You can assume that $y \neq 0$. $n = \dfrac{-6y + 18}{y + 9} \div \dfrac{y^2 + y - 12}{y + 9} $
Answer: Dividing by an expression is the same as multiplying by its inverse. $n = \dfrac{-6y + 18}{y + 9} \times \dfrac{y + 9}{y^2 + y - 12} $ First factor the quadratic. $n = \dfrac{-6y + 18}{y + 9} \times \dfrac{y + 9}{(y - 3)(y + 4)} $ Then factor out any other terms. $n = \dfrac{-6(y - 3)}{y + 9} \times \dfrac{y + 9}{(y - 3)(y + 4)} $ Then multiply the two numerators and multiply the two denominators. $n = \dfrac{ -6(y - 3) \times (y + 9) } { (y + 9) \times (y - 3)(y + 4) } $ $n = \dfrac{ -6(y - 3)(y + 9)}{ (y + 9)(y - 3)(y + 4)} $ Notice that $(y + 9)$ and $(y - 3)$ appear in both the numerator and denominator so we can cancel them. $n = \dfrac{ -6\cancel{(y - 3)}(y + 9)}{ (y + 9)\cancel{(y - 3)}(y + 4)} $ We are dividing by $y - 3$ , so $y - 3 \neq 0$ Therefore, $y \neq 3$ $n = \dfrac{ -6\cancel{(y - 3)}\cancel{(y + 9)}}{ \cancel{(y + 9)}\cancel{(y - 3)}(y + 4)} $ We are dividing by $y + 9$ , so $y + 9 \neq 0$ Therefore, $y \neq -9$ $n = \dfrac{-6}{y + 4} ; \space y \neq 3 ; \space y \neq -9 $